You want to find
P(1000 < X < 3000)
where X is normally distributed with mean 1751 and standard deviation 421. Transform X to Z, so that it follows the standard normal distribution with mean 0 and standard deviation 1 using the relation
X = 1751 + 421Z ==> Z = (X - 1751)/421
Then
P(1000 < X < 3000) = P((1000 - 1751)/421 < (X - 1751)/421 < (3000 - 1751)/421)
… ≈ P(-1.783 < Z < 2.967)
… ≈ P(Z < 2.967) - P(Z < -1.783)
… ≈ 0.9985 - 0.0373
… ≈ 0.9612
so that approximately 96.1% of the students fall in this income range.
3H2 +N2 ----> 2NH3
3 mol 2 mol
21.0 mol x mol
x=21.0*2/3=14.0 mol NH3
V = 36 pi centimeters cubed
Answer:
6x^3 − 17x + 10
Step-by-step explanation:
12x3 - 3(2x3 + 4x -1) - 5x + 7 Simplifed is 6x^3 − 17x + 10